# Simple Nim game with equal-sized piles.

Consider the standard Nim game, i.e. you can take as many coins as you want from a single pile, you should take at least one coin and you can't take coins from two or more different piles at the same time and last player to take a coin wins.

I am looking for a proof of the following statement:

Prove that if the number of piles are even and each pile contains two coins, the starting player loses.

The proof should be targeting an audience of people without previous exposure to combinatorial game theory or the Nim game jargon (balanced position, Nim sums etc.), and knows basic game theory. In short it should be a proof using simple (or strong) induction, nothing else.

Thanks.

• Note: The statement is true now, I accidentally typed odd instead of even. – firemind Apr 18 '12 at 1:30
• I have found that a description of the "Tweedledum-Tweedledee" approach to such positions of the NIM-game makes sense to a relatively general audience. I tried it out recently to an audience of mathematically talented junior high school freshmen. They were the finalists of a local math contest. I talked about NIM as a way of passing time, while my colleagues were busily grading their solutions. – Jyrki Lahtonen Apr 18 '12 at 13:14

Added: if you want to show strong induction, I would use the proposition that two equal-sized piles is a second player win. Base case is $(0,0)$, then if proved up to $(n,n)$ the first move from $(n+1,n+1)$ is to $(n+1,m)$ with $m \le n$ and the second player can move to $(m,m)$ which we know is a win. The problem with an even number of piles of $2$ for this purpose is you really have to prove for an even number of $2$'s plus and even number of $1$'s. This makes a two dimensional induction-for no piles of $2$ you have a win with no piles of $1$, with two piles of $1$, etc. Then you can induct on the (even) number of piles of $2$.